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Math Help - Lamina

  1. #1
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    Help Please

    Any help on this problem would be greatly appreciated. It deals with multiple integration:

    Find the centre of mass (x,y) of the lamina which occupies the region:
    36 less than or equal to (x^2) + (y^2) less than or equal to 49
    y greater or equal to 0
    if the density at any point is proportional to the distance from the origin.

    Thanks in advance.

    Mike
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  2. #2
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    Quote Originally Posted by JaysFan31
    Any help on this problem would be greatly appreciated. It deals with multiple integration:

    Find the centre of mass (x,y) of the lamina which occupies the region:
    36 less than or equal to (x^2) + (y^2) less than or equal to 49
    y greater or equal to 0
    if the density at any point is proportional to the distance from the origin.

    Thanks in advance.

    Mike
    Below is the diamgram of the region.
    In order to solve this problem you need to find the countinous density function. Which you say is proportional to the distance. I take this to mean that you are saying directly proportional right? Then you have,
    \frac{\mbox{density}}{\mbox{distance}}=\kappa for some constant of proportionality.
    In mathematical terms,
    \frac{\delta(x,y)}{\sqrt{x^2+y^2}}=\kappa
    Thus,
    \delta(x,y)=\kappa \sqrt{x^2+y^2}

    In order to find the center of mass you first need to find the mass. Which in this case is,
    \int_R \int \delta(x,y) dA
    Thus,
    \int_R \int \kappa \sqrt{x^2+y^2}dA
    I believe it is easier by converting to polar cordinates.
    Thus,
    \int_0^{2\pi} \int_6^7 \kappa r^2 dr \, d\theta
    Thus,
    \int_0^{2\pi} \frac{1}{3}\kappa r^3 |_6^7= 84.\bar 6\pi \kappa --->Mass
    ----------
    I will continue later.
    Attached Thumbnails Attached Thumbnails Lamina-picture6.gif  
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  3. #3
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    Quote Originally Posted by JaysFan31
    Any help on this problem would be greatly appreciated. It deals with multiple integration:

    Find the centre of mass (x,y) of the lamina which occupies the region:
    36 less than or equal to (x^2) + (y^2) less than or equal to 49
    y greater or equal to 0
    if the density at any point is proportional to the distance from the origin.

    Thanks in advance.

    Mike
    I will rewrite the definition of the region more clearly for you:

    <br />
36 \le x^2+y^2 \le 49, \ \ y \ge 0<br />
,

    which is the top half of the shaded region in hacker's diagram.

    RonL

    (If the region had corresponded to hackers shaded region the centre
    of mass would of course have been at (0,0) by symmetry)
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  4. #4
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    Quote Originally Posted by JaysFan31
    Any help on this problem would be greatly appreciated. It deals with multiple integration:

    Find the centre of mass (x,y) of the lamina which occupies the region:
    36 less than or equal to (x^2) + (y^2) less than or equal to 49
    y greater or equal to 0
    if the density at any point is proportional to the distance from the origin.

    Thanks in advance.

    Mike
    Let the density \rho(r)=\kappa r, theb=n the mass of the
    lamina is:

    <br />
M=\int_{\theta=0}^{\pi} \int_{r=6}^7 \rho(r)\ r\ dr d\theta =<br />
\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^2dr\ d\theta<br />

    Now the centre of mass is:

    <br />
\bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA<br />

    where \bold{r}=r(\cos(\theta),\sin(\theta)).

    So:

    <br />
\bold{R}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7  \kappa\ r^3\ (\cos(\theta),\sin(\theta)) dr\ d\theta<br />
.

    You should be able to complete these integrals yourself.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlank

    Now the centre of mass is:

    <br />
\bold{R}=\frac{1}{M} \int_A \rho(r)\ \bold{r} dA<br />
    Is that a line integral? You can calculate center of mass with line integral?
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  6. #6
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    Quote Originally Posted by ThePerfectHacker
    Is that a line integral? You can calculate center of mass with line integral?
    It's an area integral.
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  7. #7
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    OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

    Thanks.

    Mike
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  8. #8
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    Quote Originally Posted by JaysFan31
    OK. I was able to get the centre of mass for x (it equals zero). How does the centre of mass for y differ? It does not equal zero. What changes in the integration process?

    Thanks.

    Mike
    What changes is that \sin(\theta) is positive over the entire
    range of integration (that is (0,\pi)), so the integral over \theta
    does not vanish as it does for the \cos(\theta) in the integral
    for the x coordinate.

    RonL
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  9. #9
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    I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

    Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?
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  10. #10
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    Quote Originally Posted by JaysFan31
    I can't figure out the integral for the y-coordinate. Quick assistance would be greatly appreciated.

    Is it the integration of (r^3)sin(theta)sin(theta) drdtheta with the same bounds?
    You want:

    <br />
\bold{R_y}=\frac{1}{M}\int_{\theta=0}^{\pi} \int_{r=6}^7 \kappa\ r^3\ \sin(\theta)\ dr\ d\theta<br />

    RonL
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